22-Elec-A3 Signals and Communications · December 2019

Question 1 of 5: Full-Wave Rectifier — Fourier Series, Band-Pass Filtering and Power

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Full-Wave Rectifier — Fourier Series, Band-Pass Filtering and Power (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The fundamental frequency and real Fourier series of the rectified wave, the filter output $z(t)$, and the average powers $P_y$ and $P_z$.

Approach. Recognise that the magnitude operation halves the period, expand the rectified cosine as an even (cosine-only) Fourier series, pass the resulting line spectrum through the stated passband, and obtain both powers from Parseval's theorem.

1. Establish the fundamental frequency of the rectified wave. The input has period $T_0 = 1/f_0$, but $|\cos\theta|$ satisfies $|\cos(\theta+\pi)| = |\cos\theta|$, so the output repeats after a half-period of the input: $$T_y = \tfrac{1}{2}T_0 = \frac{1}{2 f_0}\qquad\Longrightarrow\qquad f_y = 2 f_0 = 2(120) $$ Rectification is therefore a frequency doubler, and the fundamental of the output is $$\boxed{f_y = 240\ \text{Hz}}$$
2. Set up the Fourier integral for the rectified cosine. The signal $y(t) = A|\cos(2\pi f_0 t)|$ is a real, even function of $t$, so every sine coefficient vanishes and the real series contains cosines only: $$y(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos\left(2\pi n f_y t\right),\qquad a_0 = \frac{1}{T_y}\int_{-T_y/2}^{T_y/2} y(t)\,dt$$ Over one output period the signal is simply $A\cos(2\pi f_0 t)$ for $|t| \le T_0/4$.
3. Evaluate the DC term. Substituting and integrating, $$a_0 = 2f_0\int_{-1/(4f_0)}^{1/(4f_0)} A\cos(2\pi f_0 t)\,dt = \frac{2A}{\pi}$$ With $A = 4$ V this gives $a_0 = 8/\pi = 2.546$ V — the familiar average value of a full-wave rectified sinusoid, $2/\pi \approx 63.7\%$ of the peak.
4. Evaluate the harmonic coefficients. Applying the same integral against $\cos(2\pi n f_y t) = \cos(4\pi n f_0 t)$ and using the product-to-sum identity gives the standard closed form $$a_n = \frac{2}{T_y}\int_{-T_y/2}^{T_y/2} y(t)\cos(2\pi n f_y t)\,dt = \frac{4A}{\pi}\cdot\frac{(-1)^{n+1}}{4n^2-1}$$ Note that the coefficients decay as $1/n^2$, which is the signature of a waveform that is continuous but has corners (a slope discontinuity) at each zero crossing.
5. Write the complete real Fourier series. Collecting the DC and harmonic terms with $A = 4$ V and $f_y = 240$ Hz, $$\boxed{\;y(t) = \frac{8}{\pi} + \frac{16}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{4n^2-1}\cos\!\left(2\pi\,(240n)\,t\right)\ \text{V}\;}$$ Numerically, the first few terms are

   $n$	Frequency (Hz)	$a_n$ (V)
   0 (DC)	0	+2.546
   1	240	+1.698
   2	480	−0.340
   3	720	+0.146
   4	960	−0.081

   so that $y(t) = 2.546 + 1.698\cos(480\pi t) - 0.340\cos(960\pi t) + 0.146\cos(1440\pi t) - \cdots$

6. Identify the filter passband and select the surviving line. An ideal band-pass filter of centre frequency 500 Hz and bandwidth 400 Hz passes $$f_{\text{BP}} \pm \frac{B}{2} = 500 \pm 200 \quad\Longrightarrow\quad 300\ \text{Hz} \le f \le 700\ \text{Hz}$$ Comparing against the harmonic list: 240 Hz falls below the lower edge, 720 Hz falls above the upper edge, and only the second harmonic at 480 Hz lies inside the band. The DC term is likewise rejected.
7. Write the filter output. Only the $n=2$ term survives, carried through with its own (negative) coefficient: $$\boxed{\;z(t) = -\frac{16}{15\pi}\cos\!\left(2\pi (480)t\right) = -0.3395\cos(960\pi t)\ \text{V}\;}$$ The minus sign is a genuine $180^\circ$ phase reversal, not a bookkeeping artefact: it may equivalently be written $z(t) = 0.3395\cos(960\pi t + \pi)$ V.
8. Compute the average power of $y(t)$. Because $|x(t)|^2 = x^2(t)$, rectification does not change the instantaneous power at all, so $$P_y = \overline{y^2(t)} = \overline{x^2(t)} = \frac{A^2}{2} = \frac{4^2}{2}$$ $$\boxed{P_y = 8\ \text{W (into 1}\,\Omega)}$$ Parseval's theorem provides an independent check: $P_y = a_0^2 + \tfrac{1}{2}\sum a_n^2 = 6.4845 + 1.4415 + 0.0576 + \cdots \to 8.000$ W, which converges to the same value.
9. Compute the average power of $z(t)$. A single sinusoid of amplitude $|a_2|$ carries half its squared amplitude: $$P_z = \tfrac{1}{2}a_2^2 = \tfrac{1}{2}(0.3395)^2$$ $$\boxed{P_z = 0.0576\ \text{W}}$$ That is only 0.72 % of the input power — the rectifier concentrates almost all of its output energy in the DC term and the first harmonic, both of which this filter rejects.